Question

(Sec²30+cosec ² 45) (2cos 60+sin 90+tan 45

Answer 1

Step-by-step explanation:

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Answer 2

Topic:-

Trigonometry

Question:-

(Sec²30+cosec ² 45) (2cos 60+sin 90+tan 45)

Solution:-

We know that

${sec}^{2} 30 = \dfrac{4}{3} \\ \\ {csc}^{2} 45 = 2 \\ \\ cos6 0 = \dfrac{1}{2} \\ \\ sin90 = 1 \\ \\ tan45 = 1$

$= (\dfrac{4}{3}+2)(2×\dfrac{1}{2} +1+1)$

$=(\dfrac{\cancel{10}}{\cancel{2}})(\cancel{2}×\dfrac{1}{\cancel{2}}+2)$

$=(5)×(3)$

$=15$

(Sec²30+cosec ² 45) (2cos 60+sin 90+tan 45)=15

Answer:-

15

More Information:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$